June 2023 Paper 1 Q14
14 Solve \(\quad \dfrac{8}{x} + 3x \leqslant 10 \quad\) where \(x\) is positive. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3x^2 - 10x + 8\) or \(-3x^2 + 10x - 8\) | M1 | |
| Factorising \((3x - 4)(x - 2)\) or completing the square \(\left(x - \dfrac{5}{3}\right)^2 - \dfrac{1}{9}\) or \(3\left(x - \dfrac{5}{3}\right)^2 - \dfrac{1}{3}\) or formula \(\dfrac{10 \pm \sqrt{4}}{6}\) | M1dep | \((3x - 4)(3x - 6)\) is acceptable for M1dep here |
| \(\dfrac{4}{3}\) and 2 | M1 dep | oe – such as \(1.\dot{3}\) or \(1\dfrac{1}{3}\) or \(\dfrac{4}{3}\) or \(\dfrac{8}{6}\) can be seen in any format or inequality (correct or incorrect) eg \(\dfrac{4}{3} \gt x\) |
| \(\dfrac{4}{3} \leqslant x \leqslant 2\) | A1 | oe – could be written as two separate inequalities but do not accept \(\dfrac{8}{6}\) at this stage as an oe \(\dfrac{4}{3} \leqslant x\) and \(x \leqslant 2\) must see the word ‘and’ in this format incorrect inequality signs is A0 |
Additional guidance
| Candidates may presume that there may be a negative solution and reverse the inequality They may however work this through and then realise it is giving the incorrect values when they test it. They can then correct themselves to still gain full marks | M3 A0 M3 A1 |
| Students attempting T&I to get the solutions would need to get both solutions to get M3. If they then put that into the correct inequality it will be M3 A1 | M3 A0 or M3 A1 |